Ok. I have managed to convince myself that there are no other curves. Basically because the cohomology ring of $M$ is given by $$H^*(M;\mathbb{Z})=\mathbb{Z}[\alpha,\beta]/(\alpha^3,\beta^3,\alpha \beta, \alpha^2 + \beta^2),$$ where $\alpha$ and $\beta$ are the poincare duals of the corresponding $P^1$'s, and so if $u$ is a k-covered sphere in the class $2E$, with $k\geq 2$, and $A$ is the homology class of the underlying simple sphere, we can write $A=a PD\alpha + b PD\beta$ for some integers $a,b$ and so $$kA=2E=2PD \alpha$$ implies $b=0$, $a=1$ and $k=2$, so that $A=E$ and hence the underlying simple sphere is $E$ itself, by the uniqueness of spheres mentioned above. I.e $u$ is a degree 2 cover of $E$.
Now, I am still trying to convince myself of the much more basic fact that a degree 2 map is determined by its two multiplicity 2 branch points, up to reparametrizaton. I mean, if I have a map with branch points 0 and $\infty$ say, and other with two different ones, can I not find an automorphism of the sphere by which they differ, preserving branch points? Maybe this is obviously false and someone should slap me in the face...
Also, say I fixed my two branch points, 0 and $\infty$, and I look at the degree two maps with these as branch points. Then they are polynomials of the form $az^2+bz$. But I can precompose with automorphisms preserving $0$ and $\infty$ or swapping them around, i.e maps $z \mapsto \lambda z$ and $z \mapsto \gamma/z$, and so I get a 4 dimensional space of maps quotiented out by a 4 dimensional space of automorphims, i.e a 0 dimensional thingy (?). Hopefully this is how it works...
Btw, I dont know how to comment below you... sorry for that.